How do you differentiate g(x) = (x^2+1) (x^2-3x) using the product rule?

1 Answer
Jan 4, 2016

g'(x)=4x^3-6x^2+2x-2

Explanation:

g(x)=(x^2+1)(x^2-2x)

Product rule :d/dx(uv)=(du)/dxv+u(dv)/dx

u=(x^2+1)

du/dx=2x

v=x^2-2x

dv/dx=2x=2

d/dx(x^2+1)(x^2-2x)=(du)/dxv+u(du)/dx

=2x(x^2-2x)+(x^2+1)(2x-2)

=2x^3-4x^2+2x^3-2x^2+2x-2

=4x^3-6x^2+2x-2